Two stars of masses 3 x 10³¹ kg each and at distance 2 x 10¹¹m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is $\mathrm{x}\times {10}^5{\mathrm{ms}}^{-1},\left(\mathrm{G}=6.67\times {10}^{-11}\mathrm{N}-{\mathrm{m}}^2{\mathrm{kg}}^{-2}\right)$ then the value of $x$ is

Let us assume that stars are moving in x,y- plane with origin as their centre of mass as shown in the figure below

According to question
Mass of each star, M = 3 X 10³¹kg
And diameter of circle,${\text{2R=2×10}}^{11}\mathrm{m}$
$\Rightarrow \mathrm{R}={10}^{11}\mathrm{m}$
Potential energy of meteorite at O, origin $\hat{\mathrm{j}}$ is,
${\mathrm{U}}_{\mathrm{total}}=-\frac{2\mathrm{GMm}}{\mathrm{r}}$
If v is the velocity of meteoroid at O then Kinetic energy K of the meteoroid is $\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$
To escape from this dual star system, total mechanical energy of the meteoroid at infinite distance from starts must be at least zero. By conservation of energy, we have
$\begin{array}{l}\frac{1}{2}{\mathrm{mv}}^2-\frac{2\mathrm{GMm}}{\mathrm{R}}=0\\ \Rightarrow \\ {\mathrm{v}}^2=\frac{4\mathrm{GM}}{\mathrm{R}}=\frac{4\times 6.67\times {10}^{-11}\times 3\times {10}^{31}}{{10}^{11}}\\ \Rightarrow \mathrm{v}=2.83\times {10}^5\mathrm{m}/\mathrm{s}\end{array}$